A buffer with a pH of 3.93 contains 0.17 M of sodium benzoate and 0.32 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.054 mol of HCl to a final volume of 1.5 L?
millimols base = 1,500*0.17 = 255
mmols acid = 1,500*0.32 = 480
add 54 mmols HCl
.........Bz^- + H^+ ==> HBz
I.......255......0.......480
add.............54...............
C.......-54....-54......+54
E.......201.....0.......534
Plug these E line values into the HH equation and solve for pH, then convert to (H^+). I get pH of about 3.8 or so.
To determine the concentration of [H+] in the solution after the addition of HCl, we need to apply the principles of acid-base reactions and the Henderson-Hasselbalch equation.
First, let's write out the chemical equation for the dissociation of benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa) in water:
C6H5COOH ⇌ C6H5COO- + H+
C6H5COONa → C6H5COO- + Na+
Since the buffer contains both benzoic acid and sodium benzoate, we can assume that the initial concentration of the [C6H5COO-] and [C6H5COOH] are 0.17 M and 0.32 M, respectively.
Now, we need to calculate the moles of HCl added to the solution. Given that the initial volume is 1.5 L and the amount of HCl added is 0.054 mol, the final concentration of HCl can be determined:
0.054 mol / 1.5 L = 0.036 M
Next, we will consider the acid-base reaction between HCl and the buffer components. Since benzoic acid is a weak acid while HCl is a strong acid, we can assume that the benzoic acid will primarily react with the added HCl.
The reaction can be represented as follows:
C6H5COOH + HCl ⇌ C6H5COO- + H2O
Using the stoichiometry of the reaction, we can determine the change in concentration of benzoic acid and C6H5COO-:
0.036 M (change in HCl concentration) × 1 (volume ratio benzoic acid/HCl) = 0.036 M (change in benzoic acid concentration)
0.036 M (change in HCl concentration) × 1 (volume ratio C6H5COO-/HCl) = 0.036 M (change in C6H5COO- concentration)
Now, we can calculate the final concentrations of benzoic acid ([C6H5COOH]f) and C6H5COO- ([C6H5COO-]f) in the solution:
[C6H5COOH]f = [C6H5COOH]i - (change in benzoic acid concentration) = 0.32 M - 0.036 M = 0.284 M
[C6H5COO-]f = [C6H5COO-]i + (change in C6H5COO- concentration) = 0.17 M + 0.036 M = 0.206 M
Finally, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:
pH = pKa + log([C6H5COO-]f / [C6H5COOH]f)
The pKa for benzoic acid is approximately 4.20.
pH = 4.20 + log(0.206 M / 0.284 M) = 4.20 + log(0.725) ≈ 3.93
Therefore, the concentration of [H+] in the solution after the addition of 0.054 mol of HCl to a final volume of 1.5 L is approximately equal to a pH of 3.93.